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For a one-step method applied to $y^{\prime }=\lambda y$, write $y_{n+1}=R(h\lambda )y_n$. Its absolute stability region is the set of complex $z$ where the stability function is defined and $|R(z)|\le 1$, using the bounded-iteration convention.

Explicit Euler has $R(z)=1+z$, giving the closed disk centered at $-1$ with radius one. For real $\lambda <0$, boundedness requires $h\le 2/|\lambda |$, while strict inequality gives decay. At the boundary $h=2/|\lambda |$, the numerical solution alternates with unchanged magnitude even though the exact solution decays. Some definitions use the strict region $|R(z)|<1$; stating the convention separates boundedness from asymptotic damping.

The stability function $R$ of a one-step numerical method is its amplification factor on the Dahlquist test equation $y^{\prime }=\lambda y$: one step gives $y_{n+1}=R(h\lambda )y_n$. Its absolute stability region is $\{z:|R(z)|\le 1\}$, using boundedness of repeated steps as the convention.

For explicit Euler method, $R(z)=1+z$, so the region is the closed disc centered at $-1$ with radius one. For Backward Euler method, $R(z)=1/(1-z)$, and the entire left half-plane is stable. These calculations explain why a step can be accurate locally yet generate a growing long-time solution. Multistep methods instead have several amplification roots and require a root condition.^[ref]